Integral Cake Delivery IIIIIII

For this integral, I think the best way is Weierstrass substitution which transform an integral of a rational function in sine and cosine into an integral of a rational function in t or x, no matter.

Sum up:

$\displaystyle \int F(\cos(x),\sin(x))dx \quad \underset{\text{Weierstrass substitution}}\rightarrow \quad \int G(t)dt$

Where $F$ and $G$ are rational functions ie $F=\frac{P}{Q}, \ G=\frac{R}{S}$ where $P, Q, R, S$ are polynomials ( $P$ and $Q$ have two variables).

Let's go into the maths! The Weierstrass substitution is $t=\tan(\frac{x}{2})$ . Then: $\cos(x)=2\cos^2(x/2)-1=\frac{2}{1+t^2}-1=\frac{1-t^2}{1+t^2}$

$\sin(x)=2\sin(x/2)\cos(x/2)=2t\cos^2(x/2)=\frac{2t}{1+t^2}$

$dt=\frac{1}{2}(1+t^2)dx$

ie $dx=\frac{2}{1+t^2}dt$

Plugging in our integral:

$\displaystyle \int_{0}^{π/2} \frac{dx}{(1+\cos(x))^2} \\ = \displaystyle \int_{0}^{1}\frac{2dt}{(1+t^2)(1+\frac{1-t^2}{1+t^2})^2} \\ = \displaystyle \int_{0}^{1}\frac{2(1+t^2)}{(1+t^2+1-t^2)^2}dt \\ = \displaystyle \frac{1}{2}\int_{0}^{1}(1+t^2)dt \\ =\boxed{\frac{2}{3}}$